Show that the points A(3, 5), B(6, 0), C(1, -3) and D(-2, 2) are the vertices of a square ABCD.

Explanation

Let A (3, 5), B (6, 0), C(1, -3) and D(-2, 2) be the angular points of a quadrilateral ABCD. Join AC and BD.
Now, `AB = sqrt((6-3)^(2) + (0-5)^(2))`
`=sqrt(3^(2) + (-5)^(2))`
`=sqrt(9+25) = sqrt(34)` units,
`BC = sqrt((1-6)^(2) + (-3-0)^(2)) = sqrt((-5)^(2)+ (-3)^(2))`
`sqrt(25+9) = sqrt(34)` units,
`CD = sqrt((-2-1)^(2) + (2+3)^(2)) = sqrt((-3)^(2) + 5^(2))`
`=sqrt(9+25) = sqrt(34)` units,
`"and"` DA = sqrt((3+2)^(2) + (5-2)^(2)) = sqrt(5^(2) +3^(2))`
`= sqrt(25+9) = sqrt(34)` units.
Thus, AB = BC = CD = DA.
Diag. `AC = sqrt((1-3)^(2) + (-3-5)^(2)) = sqrt((-2)^(2) + (-8)^(2))`
`=sqrt(4+64) = sqrt(68) = 2sqrt(17)` units.
Diag. `BD = sqrt((-2-6)^(2) + (2-0)^(2))`
` = sqrt((-8)^(2) + 2^(2) = sqrt(64+4)`
`=sqrt(68) = 2 sqrt(17)` units.
`therefore` diag. AC = diag. BD.
Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.
Hence, quad. ABCD is a square.